---
title: "Mechanical Energy and Its Conservation"
book: "High School Physics"
subject: physics
language: en
chapter: 18
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation
---

# Chapter 18 — Mechanical Energy and Its Conservation

A roller coaster has no engine past the first hill: every loop, every burst of speed afterwards is spent from the account opened on the way up. This chapter turns the bookkeeping into two numbers — kinetic and potential [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) — and one rule: while only [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) [works](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work), their sum does not change. [Friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) is the fee; we will learn to read it off the books.

## 18.1 Kinetic energy

**Definition 18.1 (Kinetic energy).**

A body of mass $m$ ($\mathrm{kg}$) moving at speed $v$ ($\mathrm{m}/\mathrm{s}$) carries the *kinetic energy*

$$
E_k = \tfrac12\, m v^2 .
$$

[Units](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) check: $\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}^{2} = \mathrm{N}\,\mathrm{m} = \mathrm{J}$ — [joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work), like every [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) ([Chapter 9](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#ch-g10-energy-conservation)).

**Example 18.2 (Orders of magnitude).**

- a pedestrian, $70\,\mathrm{kg}$ at $1.4\,\mathrm{m}/\mathrm{s}$ : $E_k = \tfrac12 \times 70 \times 1.4^2 \approx 69\,\mathrm{J}$ ;
- a car, $1300\,\mathrm{kg}$ at $130\,\mathrm{km}/\mathrm{h}$ $= 36.1\,\mathrm{m}/\mathrm{s}$ : $E_k \approx 8.5 \times 10^{5}\,\mathrm{J}$ ;
- a high-speed train, $m = 3.85 \times 10^{5}\,\mathrm{kg}$ , at $320\,\mathrm{km}/\mathrm{h}$ $= 88.9\,\mathrm{m}/\mathrm{s}$ : $E_k \approx 1.5 \times 10^{9}\,\mathrm{J}$ ;
- a rifle bullet, $8.0\,\mathrm{g}$ at $800\,\mathrm{m}/\mathrm{s}$ : $E_k \approx 2.6\,\mathrm{kJ}$ .

Seven orders of magnitude separate the stroll from the train. And $E_k$ grows like the *square* of the speed: at $130\,\mathrm{km}/\mathrm{h}$ the car carries $4$ times the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) it has at $65\,\mathrm{km}/\mathrm{h}$ — four times what its brakes must remove.

## 18.2 The kinetic-energy theorem

**Theorem 18.3 (Kinetic-energy theorem).**

For a body of mass $m$ moving along a straight line from $A$ to $B$, the change of [kinetic energy](#def-g11-mechanical-energy-kinetic) equals the total [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) ([Chapter 17](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#ch-g11-work-of-force)) received:

$$
\Delta E_k = E_k(B) - E_k(A) = \sum W_{AB}(\vect F),
$$

the sum running over all [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) acting on the body. It extends to motion along any path, as we also admit here.

**Proof.** *Admitted at this level.* ∎

**Example 18.4 (Free fall, checked frame by frame).**

A $0.200\,\mathrm{kg}$ ball is dropped and filmed; the frames give:

| $t$ ($\mathrm{s}$) | 0.10 | 0.20 | 0.30 | 0.40 |
| --- | --- | --- | --- | --- |
| fallen height $h$ ($\mathrm{m}$) | 0.049 | 0.196 | 0.441 | 0.785 |
| speed $v$ ($\mathrm{m}/\mathrm{s}$) | 0.98 | 1.96 | 2.94 | 3.92 |
| $E_k = \tfrac12 mv^2$ ($\mathrm{J}$) | 0.096 | 0.384 | 0.864 | 1.54 |
| $W = mgh$ ($\mathrm{J}$) | 0.096 | 0.385 | 0.865 | 1.54 |

Starting from $E_k = 0$, at every frame $E_k$ matches the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) $mgh$ of the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight): $\Delta E_k = W(\vect P)$, i.e. $v^2 = 2gh$ — the theorem, verified to the precision of the readings.

**Remark 18.5 (Where the proof lives).**

The general statement — valid along any path, not only a straight line — follows from Newton’s second law and a little calculus, both later in this book (Chapters [25](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws) and [29](https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy#ch-g12-work-and-energy)); until then the free-fall check is our warrant, and we admit the curved-path version and use it freely.

## 18.3 Gravitational potential energy

**Definition 18.6 (Gravitational potential energy).**

A body of mass $m$ at altitude $z$, measured upward from a chosen *reference level* where $z = 0$, stores the *gravitational potential energy*

$$
E_p = m g z, \qquad g = 9.81\,\mathrm{N}/\mathrm{kg}.
$$

**Remark 18.7 (Only differences matter).**

Moving from $A$ to $B$ changes $E_p$ by $mg(z_B - z_A) =
-W_{AB}(\vect P)$: the opposite of the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) of the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) ([Chapter 17](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#ch-g11-work-of-force)). The [reference level](#def-g11-mechanical-energy-potential) is a free choice — floor, sea level, table top — shifting $E_p$ everywhere by a constant that drops out of every difference. Put it where the numbers are simplest, and do not move it mid-problem.

**Example 18.8 (A book on a shelf).**

A $1.2\,\mathrm{kg}$ book sits $1.8\,\mathrm{m}$ above the floor, itself $9.0\,\mathrm{m}$ above the street. Reference at the floor: $E_p = 1.2 \times 9.81 \times 1.8 \approx 21\,\mathrm{J}$; in the street: $E_p \approx 127\,\mathrm{J}$. Falling to the floor releases $21\,\mathrm{J}$ in both accounts.

## 18.4 Mechanical energy and its conservation

**Definition 18.9 (Mechanical energy).**

The *mechanical energy* of a body is

$$
E_m = E_k + E_p = \tfrac12 m v^2 + m g z .
$$

**Theorem 18.10 (Conservation of mechanical energy).**

If the only [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) working on a body is its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) (no [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory); other [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), like the reaction of a frictionless track, perpendicular to the motion), $E_m$ is conserved: for any two points $A$, $B$ of the motion,

$$
\tfrac12 m v_A^2 + m g z_A = \tfrac12 m v_B^2 + m g z_B .
$$

**Proof.** By the [kinetic-energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) theorem, $E_k(B) - E_k(A) = W_{AB}(\vect P) =
mg(z_A - z_B) = E_p(A) - E_p(B)$: what $E_k$ gains, $E_p$ loses. ∎

**Example 18.11 (Water slide).**

A child starts at rest atop a frictionless slide of height $h = 3.2\,\mathrm{m}$. Reference at the bottom: $mgh = \tfrac12 mv^2$, so $v = \sqrt{2gh} \approx 7.9\,\mathrm{m}/\mathrm{s}$ ($29\,\mathrm{km}/\mathrm{h}$) — whatever the mass *and whatever the shape of the slide*: straight, curved or spiral, only the height drop enters.

**Example 18.12 (Pendulum).**

A pendulum bob is pulled aside until it rises $h = 12\,\mathrm{cm}$ above its lowest point, then released. At the extremes $v = 0$; at the bottom the whole $mgh$ is kinetic, $v = \sqrt{2gh} \approx 1.5\,\mathrm{m}/\mathrm{s}$; on the far side the bob climbs back to exactly $12\,\mathrm{cm}$.

![The pendulum trades E_p for E_k and back: all potential at the extremes, all kinetic at the bottom, the same rise h twice.](https://one-course.com/images/onecourse/chapters/physics-2/g11-mechanical-energy/fig-8240e3c22564.svg)

*The pendulum trades $E_p$ for $E_k$ and back: all potential at the extremes, all kinetic at the bottom, the same rise $h$ twice.*

![Energy bar chart of a frictionless coaster: at every position the E_p and E_k bars stack to the same total E_m (dashed line).](https://one-course.com/images/onecourse/chapters/physics-2/g11-mechanical-energy/fig-e1246bdcf5fd.svg)

*[Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) bar chart of a frictionless coaster: at every position the $E_p$ and $E_k$ bars stack to the same total $E_m$ (dashed line).*

**Example 18.13 (Ski jump landing speed).**

A ski jumper leaves the takeoff at $v_A = 25\,\mathrm{m}/\mathrm{s}$ and lands $\Delta z = 40\,\mathrm{m}$ lower. Neglecting air resistance, $v_B = \sqrt{v_A^2 + 2g\,\Delta z} = \sqrt{625 + 785} \approx
38\,\mathrm{m}/\mathrm{s}$ — no knowledge of the flight path needed.

![Ski jump: whatever the curve flown from A to B, only the drop z enters the energy balance.](https://one-course.com/images/onecourse/chapters/physics-2/g11-mechanical-energy/fig-77b4169c48e1.svg)

*Ski jump: whatever the curve flown from $A$ to $B$, only the drop $\Delta z$ enters the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) balance.*

## 18.5 Dissipation by friction

**Definition 18.14 (Dissipation).**

[Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) leaving the mechanical account is said to be *dissipated*: it reappears as [thermal energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-forms) in the rubbing surfaces and the air. Total [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) is conserved ([Chapter 9](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#ch-g10-energy-conservation)); [mechanical energy](#def-g11-mechanical-energy-em) alone is not.

**Proposition 18.15 (The deficit measures friction).**

If a [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\vect f$ also [works](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) along the motion,

$$
\Delta E_m = E_m(B) - E_m(A) = W_{AB}(\vect f) < 0:
$$

the loss of [mechanical energy](#def-g11-mechanical-energy-em) equals the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) of [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory).

**Proof.** The [kinetic-energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) theorem now reads $\Delta E_k = W_{AB}(\vect P) +
W_{AB}(\vect f) = -\Delta E_p + W_{AB}(\vect f)$; move $\Delta E_p$ across. [Friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) opposes the motion, so its [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) is negative. ∎

**Example 18.16 (Toboggan audit).**

A child and sled ($m = 40\,\mathrm{kg}$) start at rest, drop $h = 5.0\,\mathrm{m}$ along $12\,\mathrm{m}$ of snow, and arrive at $7.0\,\mathrm{m}/\mathrm{s}$ instead of the frictionless $\sqrt{2gh} \approx 9.9\,\mathrm{m}/\mathrm{s}$. Reference at the bottom: $\Delta E_m = \tfrac12 \times 40 \times 7.0^2 - 40 \times 9.81
\times 5.0 = 980 - 1962 = -982\,\mathrm{J}$. So $W(\vect f) =
-982\,\mathrm{J}$ over $12\,\mathrm{m}$: an average [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $f = 982/12 \approx 82\,\mathrm{N}$, and $982\,\mathrm{J}$ of warmth in the runners and the snow.

![E_m along a motion: level without friction, decreasing with it — the drop at each instant is the heat produced so far.](https://one-course.com/images/onecourse/chapters/physics-2/g11-mechanical-energy/fig-b748fbf20438.svg)

*$E_m$ along a motion: level without [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory), decreasing with it — the drop at each instant is the heat produced so far.*

**Method 18.17 (Energy bookkeeping).**

1. Choose a [reference level](#def-g11-mechanical-energy-potential) and keep it; name $A$ (data), $B$ (question).
2. Write $E_m = \tfrac12 mv^2 + mgz$ at $A$ and at $B$ .
3. No [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) : $E_m(A) = E_m(B)$ ; solve — typically $v_B = \sqrt{v_A^2 + 2g(z_A - z_B)}$ .
4. [Friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) : $E_m(B) - E_m(A) = -f\ell$ ( $\ell$ = path length); solve for the unknown.
5. Check: the $E_k$ / $E_p$ bars stack to $E_m$ , level unless [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) .

## 18.6 Exercises

**Exercise 18.1 ★.**

Compute the [kinetic energy](#def-g11-mechanical-energy-kinetic) of (a) a $90\,\mathrm{kg}$ scooter and rider at $25\,\mathrm{km}/\mathrm{h}$; (b) a $1200\,\mathrm{kg}$ car at $130\,\mathrm{km}/\mathrm{h}$. (c) By what factor does $E_k$ change when the speed doubles? Triples?

**Solution of Exercise 18.1.**

(a) $v = 6.94\,\mathrm{m}/\mathrm{s}$: $E_k = \tfrac12 \times 90 \times 6.94^2 \approx 2.2 \times 10^{3}\,\mathrm{J}$. (b) $v = 36.1\,\mathrm{m}/\mathrm{s}$: $E_k = \tfrac12 \times 1200 \times 36.1^2 \approx 7.8 \times 10^{5}\,\mathrm{J}$. (c) $\times 4$; $\times 9$ — the square.

**Exercise 18.2 ★.**

A $75\,\mathrm{kg}$ climber gains $850\,\mathrm{m}$ of altitude. Compute the gain of potential [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy). Does it depend on the [reference level](#def-g11-mechanical-energy-potential)? On the route?

**Solution of Exercise 18.2.**

$\Delta E_p = 75 \times 9.81 \times 850 \approx 6.3 \times 10^{5}\,\mathrm{J}$. No: the reference shifts both values equally. No: only the height difference enters.

**Exercise 18.3 ★.**

A flowerpot falls $20\,\mathrm{m}$ from a balcony. Neglecting air resistance, find its ground speed in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$. Where did the mass go?

**Solution of Exercise 18.3.**

$v = \sqrt{2 \times 9.81 \times 20} = 19.8\,\mathrm{m}/\mathrm{s} \approx
71\,\mathrm{km}/\mathrm{h}$. $mgh = \tfrac12 mv^2$: the mass cancels.

**Exercise 18.4 ★.**

A $1200\,\mathrm{kg}$ car brakes from $25\,\mathrm{m}/\mathrm{s}$ to rest in $50\,\mathrm{m}$. Use the [kinetic-energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) theorem to find the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) of the brakes, then the average braking [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force).

**Solution of Exercise 18.4.**

$W = \Delta E_k = 0 - \tfrac12 \times 1200 \times 25^2 =
-3.75 \times 10^{5}\,\mathrm{J}$; $F = 3.75 \times 10^{5}/50 = 7.5 \times 10^{3}\,\mathrm{N}$.

**Exercise 18.5 ★.**

A pendulum bob is released $8.0\,\mathrm{cm}$ above its lowest point. Find its speed at the bottom and the height reached on the other side.

**Solution of Exercise 18.5.**

$v = \sqrt{2 \times 9.81 \times 0.080} \approx 1.3\,\mathrm{m}/\mathrm{s}$; it rises back to $8.0\,\mathrm{cm}$ (conservation).

**Exercise 18.6 ★★.**

Using [Example 18.2](#ex-g11-mechanical-energy-orders): by what factor does the train’s [kinetic energy](#def-g11-mechanical-energy-kinetic) exceed the bullet’s? To what height could its own $E_k$ lift the train ($h = v^2/2g$)?

**Solution of Exercise 18.6.**

$1.5 \times 10^{9}/2.6 \times 10^{3} \approx 5.9 \times 10^{5}$: about six hundred thousand bullets. $h = 88.9^2/(2 \times 9.81) \approx 4.0 \times 10^{2}\,\mathrm{m}$ — its own speed could lift the train $400$ meters.

**Exercise 18.7 ★★.**

Two frictionless slides drop from the same $3.0\,\mathrm{m}$ platform, one straight and steep, one long and winding. Compare the arrival speeds (compute them) and the arrival times (no computation).

**Solution of Exercise 18.7.**

Same speed $v = \sqrt{2 \times 9.81 \times 3.0} \approx 7.7\,\mathrm{m}/\mathrm{s}$ (only the drop enters). The steep slide wins on time: it reaches high speed sooner and its path is shorter.

**Exercise 18.8 ★★.**

A ball is thrown straight up at $12\,\mathrm{m}/\mathrm{s}$. Find the maximum height reached, then the speed at $4.0\,\mathrm{m}$ — on the way up and on the way down.

**Solution of Exercise 18.8.**

$h = v^2/2g = 144/19.62 \approx 7.3\,\mathrm{m}$. At $4.0\,\mathrm{m}$: $v = \sqrt{144 - 2 \times 9.81 \times 4.0} \approx 8.1\,\mathrm{m}/\mathrm{s}$ — the same both ways: $E_m$ depends on height, not on direction of travel.

**Exercise 18.9 ★★.**

A ski jumper leaves the takeoff at $23\,\mathrm{m}/\mathrm{s}$ and lands $35\,\mathrm{m}$ lower. Predict the landing speed without air resistance; is the real value larger or smaller, and why?

**Solution of Exercise 18.9.**

$v = \sqrt{23^2 + 2 \times 9.81 \times 35} = \sqrt{1216} \approx
35\,\mathrm{m}/\mathrm{s}$. Smaller in reality: air resistance dissipates part of $E_m$.

**Exercise 18.10 ★★.**

A $45\,\mathrm{kg}$ sled starts at rest, drops $6.0\,\mathrm{m}$ along $15\,\mathrm{m}$ of slope, arriving at $8.0\,\mathrm{m}/\mathrm{s}$. Find the [mechanical energy](#def-g11-mechanical-energy-em) lost, then the average [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force). Where did the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) go?

**Solution of Exercise 18.10.**

$\Delta E_m = \tfrac12 \times 45 \times 8.0^2 - 45 \times 9.81 \times
6.0 = 1440 - 2649 \approx -1.2 \times 10^{3}\,\mathrm{J}$; $f = 1209/15 \approx 81\,\mathrm{N}$. Into heat, in the runners and the snow.

**Exercise 18.11 ★★.**

Galileo’s peg: a pendulum released $15\,\mathrm{cm}$ above its lowest point meets a peg blocking the upper half of the string at the vertical. Find the speed at the bottom and the rise beyond the peg; what does the peg change, and what not?

**Solution of Exercise 18.11.**

$v = \sqrt{2 \times 9.81 \times 0.15} \approx 1.7\,\mathrm{m}/\mathrm{s}$; it rises to $15\,\mathrm{cm}$ again. The peg changes the path (a tighter circle), not the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy): heights and speeds are untouched.

**Exercise 18.12 ★★★.**

A skateboarder drops into a half-pipe from $2.5\,\mathrm{m}$; each traverse (side to side) dissipates $10\%$ of the [mechanical energy](#def-g11-mechanical-energy-em). Find the speed at the bottom of the first descent, then — computing the successive rise heights — the first traverse ending below $1.0\,\mathrm{m}$.

**Solution of Exercise 18.12.**

$v = \sqrt{2 \times 9.81 \times 2.5} = 7.0\,\mathrm{m}/\mathrm{s}$. Rise heights scale with $E_m$, so $\times 0.9$ each traverse: $2.25$, $2.03$, $1.82$, $1.64$, $1.48$, $1.33$, $1.20$, $1.08$, $0.97$ — first below $1.0\,\mathrm{m}$ on the $9$th traverse.

**Exercise 18.13 ★★★.**

From a $45\,\mathrm{m}$ cliff top, a stone is thrown at $15\,\mathrm{m}/\mathrm{s}$ — upward, horizontally, or at any angle. Show that the landing speed is the same in all cases and compute it. What *does* depend on the angle?

**Solution of Exercise 18.13.**

Same initial $E_m = \tfrac12 m \times 15^2 + m \times 9.81 \times 45$ whatever the angle, so the ground speed is $v = \sqrt{225 + 2 \times 9.81 \times 45} = \sqrt{1108} \approx
33\,\mathrm{m}/\mathrm{s}$. The angle fixes the *direction* of the landing velocity, the flight time and the [range](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-strong) — not the landing speed.

**Exercise 18.14 ★★★.**

An $80\,\mathrm{kg}$ downhill skier starts at rest and descends $120\,\mathrm{m}$ of drop along an $800\,\mathrm{m}$ run, against a constant $65\,\mathrm{N}$ [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force). Find the arrival speed and the fraction of $E_m$ [dissipated](#def-g11-mechanical-energy-dissipation).

**Solution of Exercise 18.14.**

$\tfrac12 mv^2 = mgh - f\ell = 80 \times 9.81 \times 120 - 65 \times
800 = 94\,176 - 52\,000 = 4.22 \times 10^{4}\,\mathrm{J}$, so $v = \sqrt{2 \times 42\,176/80} \approx 32\,\mathrm{m}/\mathrm{s}$ ($117\,\mathrm{km}/\mathrm{h}$). Fraction [dissipated](#def-g11-mechanical-energy-dissipation): $52000/94176 \approx 55\%$.

**Exercise 18.15 ★★★.**

A pumped-storage plant lifts $V = 2.0 \times 10^{6}\,\mathrm{m}^{3}$ of water ($1000\,\mathrm{kg}$ per $\mathrm{m}^{3}$) through $300\,\mathrm{m}$. Compute the stored [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) in [joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) and in $\mathrm{kW}\,\mathrm{h}$; turbined back at $85\%$ [efficiency](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) ([Chapter 9](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#ch-g10-energy-conservation)), how many homes drawing $10\,\mathrm{kW}\,\mathrm{h}$ per day does it feed for one day?

**Solution of Exercise 18.15.**

$m = 2.0 \times 10^{9}\,\mathrm{kg}$: $E_p = 2.0 \times 10^{9} \times 9.81 \times 300 \approx 5.9 \times 10^{12}\,\mathrm{J}
= 5.9 \times 10^{12}/3.6 \times 10^{6} \approx 1.6 \times 10^{6}\,\mathrm{kW}\,\mathrm{h}$. Recovered: $0.85 \times 1.64 \times 10^{6} \approx 1.4 \times 10^{6}\,\mathrm{kW}\,\mathrm{h}$ — about $1.4 \times 10^{5}$ homes for a day.

## 18.7 Problem: Designing the loop

**Problem 18.1.**

Weekend problem — an energy audit of a roller coaster: the launch height a loop dictates, the speeds it promises, the friction confessed by two sensors, the brakes that close the account — the whole ride hanging from one number

A train of mass $m = 2.0 \times 10^{3}\,\mathrm{kg}$ is released at rest from a launch hill of height $H$ and coasts, engineless, through a vertical circular loop of radius $R = 8.0\,\mathrm{m}$ whose top sits at height $2R$. Circular-motion analysis (next year) supplies the one datum we borrow: at the loop top the train presses on the track only if $v_{\text{top}}^2 \geq gR$. Parts I and II ignore [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory).

**Part I — The hill the loop dictates.**

1. Compute the minimum safe speed at the top of the loop.
2. Ground as [reference level](#def-g11-mechanical-energy-potential) : compute the train’s [mechanical energy](#def-g11-mechanical-energy-em) at the loop top at that minimum speed.
3. Deduce $H_{\min}$ and show $H_{\min} = 5R/2$ — no mass, no $g$ .
4. The team takes $H = 24\,\mathrm{m}$ : compute the actual top-of-loop speed and the ratio $v_{\text{top}}^2/(gR)$ .
5. Where did $m$ go? Why does the design fit full and empty trains alike?

**Part II — What the height promises** ($H = 24\,\mathrm{m}$).

6. Speed at the bottom of the loop ( $z = 0$ ), in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ ?
7. Speed at the side of the loop ( $z = R$ )?
8. Compute $E_p$ and $E_k$ at launch, bottom, side and top of the loop; check that each pair sums to $mgH$ .
9. A $12\,\mathrm{m}$ camel-back bump follows the loop: speed at its crest?
10. Give the general formula for $v(z)$ ; where is the ride fastest, and slowest?

**Part III — The [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) audit.** Two sensors at the same height $z = 2.0\,\mathrm{m}$, $60\,\mathrm{m}$ of track apart, read $v_1 = 20.4\,\mathrm{m}/\mathrm{s}$ then $v_2 = 19.2\,\mathrm{m}/\mathrm{s}$.

11. Compute the change of [mechanical energy](#def-g11-mechanical-energy-em) between the sensors. What made the same-height placement clever?
12. Deduce the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) of [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) and the average [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) .
13. Where has the missing [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) gone? Name the form and the places.
14. What fraction of the train’s $E_m$ is lost per $60\,\mathrm{m}$ ? Why would the bare $H_{\min}$ be unsafe in the real, rubbing world?
15. About $90\,\mathrm{m}$ of track separate launch and loop top: estimate $E_m$ there; does the margin of question 4 still save the loop?

**Part IV — Closing the account.** The last sensor, at the entry of the straight horizontal braking run ($z = 0$), reads $18.0\,\mathrm{m}/\mathrm{s}$.

16. Compute the train’s [kinetic energy](#def-g11-mechanical-energy-kinetic) there.
17. The brakes must stop it in $d = 45\,\mathrm{m}$ : compute the required (constant) braking [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) .
18. Comfort check: compute the deceleration $a = F/m$ ; compare to $g$ .
19. How much [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) did track [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) dissipate over the whole ride? Verify that [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $+$ brakes $= mgH$ , to the [joule](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) .
20. Punchline, one sentence: which single number fixed the loop’s safety, every speed, the [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) bill and the brake [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) — and what was the price of ignoring [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) ?

**Solution of Problem 18.1.**

**1.** $v_{\text{top}} = \sqrt{gR} = \sqrt{9.81 \times 8.0}
\approx 8.9\,\mathrm{m}/\mathrm{s}$.

**2.** $E_m = mg\,(2R) + \tfrac12 m gR = 3.14 \times 10^{5} +
7.8 \times 10^{4} \approx 3.92 \times 10^{5}\,\mathrm{J}$.

**3.** $mgH_{\min} = E_m$ gives $H_{\min} = 20\,\mathrm{m}$; algebraically $H_{\min} = 2R + \frac{gR}{2g} = \frac{5R}{2}$.

**4.** $v_{\text{top}} = \sqrt{2g(H - 2R)} =
\sqrt{2 \times 9.81 \times 8.0} \approx 12.5\,\mathrm{m}/\mathrm{s}$; $v_{\text{top}}^2/(gR) = 2.0$ — twice the minimum.

**5.** Every [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) is proportional to $m$, so $m$ cancels: same speeds and same safety, full or empty.

**6.** $v = \sqrt{2 \times 9.81 \times 24} \approx 21.7\,\mathrm{m}/\mathrm{s}
\approx 78\,\mathrm{km}/\mathrm{h}$.

**7.** $v = \sqrt{2 \times 9.81 \times (24 - 8.0)} \approx
17.7\,\mathrm{m}/\mathrm{s}$.

**8.** $mgH = 4.71 \times 10^{5}\,\mathrm{J}$. $(E_p, E_k)$: launch $(4.71 \times 10^{5}, 0)$; bottom $(0, 4.71 \times 10^{5})$; side $(1.57 \times 10^{5}, 3.14 \times 10^{5})$; top $(3.14 \times 10^{5}, 1.57 \times 10^{5})$ — each pair sums to $4.71 \times 10^{5}\,\mathrm{J}$.

**9.** $v = \sqrt{2 \times 9.81 \times (24 - 12)} \approx
15.3\,\mathrm{m}/\mathrm{s}$.

**10.** $v(z) = \sqrt{2g(H - z)}$: fastest at the lowest point of the track, slowest at the top of the loop.

**11.** Same height, so $\Delta E_p = 0$ and $\Delta E_m = \Delta E_k = \tfrac12 \times 2000 \times (19.2^2 -
20.4^2) \approx -4.75 \times 10^{4}\,\mathrm{J}$: the placement makes the deficit pure kinetic, read off two speedometers.

**12.** $W(\vect f) = -4.75 \times 10^{4}\,\mathrm{J}$; $f = 4.75 \times 10^{4}/60 \approx 7.9 \times 10^{2}\,\mathrm{N}$.

**13.** [Thermal energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-forms), in the wheels, the rails and the air.

**14.** $E_m$ at the first sensor: $4.16 \times 10^{5} + 3.9 \times 10^{4} =
4.55 \times 10^{5}\,\mathrm{J}$; loss $4.75 \times 10^{4}/4.55 \times 10^{5} \approx 10\%$ per $60\,\mathrm{m}$. At $H_{\min}$ the top of the loop is reached *exactly* at $\sqrt{gR}$; any loss drops it below the safety speed.

**15.** Losing $\approx 10\%$ per $60\,\mathrm{m}$, about $15\%$ of $mgH$ ($\approx 7.1 \times 10^{4}\,\mathrm{J}$) is gone: $E_m \approx 4.00 \times 10^{5}\,\mathrm{J}$, still above the required $3.92 \times 10^{5}\,\mathrm{J}$ — safe, with only $2\%$ to spare. The margin was no luxury.

**16.** $E_k = \tfrac12 \times 2000 \times 18.0^2 =
3.24 \times 10^{5}\,\mathrm{J}$.

**17.** $F = 3.24 \times 10^{5}/45 = 7.2 \times 10^{3}\,\mathrm{N}$.

**18.** $a = 7200/2000 = 3.6\,\mathrm{m}/\mathrm{s}^{2} \approx 0.37\,g$: firm but comfortable.

**19.** Track [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory): $mgH - 3.24 \times 10^{5} = 470\,880 -
324\,000 = 1.47 \times 10^{5}\,\mathrm{J}$. Audit: $1.47 \times 10^{5} + 3.24 \times 10^{5} =
4.71 \times 10^{5} = mgH$ — every [joule](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) accounted for.

**20.** The launch height $H = 24\,\mathrm{m}$ fixed the loop’s safety, every speed $\sqrt{2g(H - z)}$, the [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) bill and the brake [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force); the price of ignoring [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) is the margin — the $4\,\mathrm{m}$ of hill that the rubbing world quietly consumes.